JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2023
- Hard
- Numerical
A square loop of side \(2.0\,cm\) is placed inside a long solenoid that has \(50\) turns per centimetre and carries a sinusoidally varying current of amplitude \(2.5\,A\) and angular frequency \(700\,rad\,s ^{-1}\). The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is \(x \times 10^{-4} V\). The value of \(x\) is \(.........\)(Take, \(\pi=\frac{22}{7}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(44\)
Step-by-step Solution
Detailed explanation
\(B _{\text {due to solenoid }}=\mu_0 nI\)
\(\Phi_{\text {through square }}=\mu_0 nI \times A \quad( A =\text { Area })\)
\(Emf =\mu_0 nA \times \frac{ dI }{ dt }\)
\(=\mu_0 n A \times I _0 \omega \cos \omega t\)
\(\text { Emf amplitude }=\mu_0 n A \times I _0 \omega\)
\(=4 \pi \times 10^{-7} \times \frac{50}{10^{-2}} \times 4 \times 10^{-4} \times 2.5 \times 700\)
\(=44 \times 10^{-4}\,V\)
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